How can I determine the point(s), if any, at which the graph of the function has a horizontal tangent line?
Given: $\displaystyle y=x^2+2x$
Any help is welcomed
hi hemi
differentiate your function
you will get dy/dx. it is the slope of the tangent.
since in the question it has asked for horizontal line,means tangent parellel to x axis.
therefore, its slope will be zero.
put dy/dx=0
and get the value of x
put this value in the function and get the value of y.
thank you
hi hemi
this is your solution
y=2*x+x^2
dy/dx=2+2*x. this is the slope of the tangent.
since tangent is parellel to x-axis therefore
dy/dx=0
=> 2+2*x=0
=> x=-1
put this value in function to get the value of y
=> y=-1
therefore, the point on the function at which tangent of the function will be horizontal is (-1,-1).
Thank You
You could have done this, by the way, without using Calculus. This is a quadratic function so its graph is a vertical parabola. A tangent line will be horizontal only at its vertex. And we can get the vertex by completing the square: $\displaystyle y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+1)^2- 1$. For x= -1 we have $\displaystyle y= 0^2+ 1= 1$. For any other value of x, x+ 1 is NOT 0 so $\displaystyle (x+ 1)^2$ is positive and y is greater than 1. (-1, 1) is the vertex.